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Finding Node Voltages Using the Node Method

by erok81
Tags: method, node, voltages
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erok81
#1
Aug31-11, 08:20 PM
P: 463
1. The problem statement, all variables and given/known data

Find the node voltages for the circuit shown.

2. Relevant equations

See attachment.

3. The attempt at a solution

I have chosen my ground node as the set of nodes in the bottom of the image and labeled all of my unknown voltages as en.

Here is what I have so far. Where G = 1/R

For the node labeled e1:

(e1-v0)G1 + e1(G3) + (e1-e2)I1 + (e1-e2)G2=0

For the node labeled e2:

(e2-e1)(-I2) + e2(G4) + (e2-e1)G2=0

I am pretty sure my problem lies in the way I am handling the currents in the node. I started plugging in numbers after I simplified it all and one of my unknown voltages went away. After that I looked at it again and knew I was doing it wrong. Most of the examples in the book don't deal with a current parallel to a resistor.

I think the way I should have done it is rather than subtracting voltages and multiplying like I am doing with the resistor nodes, is just take the current by itself.

So in my above equation e1)(-I1) would just become -I1 since I am not using ohms law for current. I think I only need current? Does that sound better?
Attached Thumbnails
circuit.png  
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gneill
#2
Aug31-11, 10:23 PM
Mentor
P: 11,683
Quote Quote by erok81 View Post
I am pretty sure my problem lies in the way I am handling the currents in the node. I started plugging in numbers after I simplified it all and one of my unknown voltages went away. After that I looked at it again and knew I was doing it wrong. Most of the examples in the book don't deal with a current parallel to a resistor.

I think the way I should have done it is rather than subtracting voltages and multiplying like I am doing with the resistor nodes, is just take the current by itself.

So in my above equation e1)(-I1) would just become -I1 since I am not using ohms law for current. I think I only need current? Does that sound better?
Yup. Much better. Voltage x current yields power, not a current, so is quite unsuitable for a KCL expression!
erok81
#3
Sep1-11, 01:26 PM
P: 463
Perfect! Thanks for the help.

It seems half the time I post problems, I get them mostly figured out just typing the post up. :)

gneill
#4
Sep1-11, 02:51 PM
Mentor
P: 11,683
Finding Node Voltages Using the Node Method

Quote Quote by erok81 View Post
It seems half the time I post problems, I get them mostly figured out just typing the post up. :)
Yup. Typing it out so that it make sense to someone else can often help one re-evaluate one's logic and assumptions.


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